If $f\left(a\right)=3,f^{\mathrm{\prime }}\left(a\right)=-2,g\left(a\right)=-1,g^{\mathrm{\prime }}\left(a\right)=4,$ then $\underset{x\to a}{lim} \frac{g\left(x\right)f\left(a\right)-g\left(a\right)f\left(x\right)}{x-a}=?$

   $\underset{x\to a}{lim} \frac{g\left(x\right)f\left(a\right)-g\left(a\right)f\left(x\right)}{x-a}.$

We add and subtract $g\left(a\right)f\left(a\right)$ in numerator

$\begin{array}{l}=\underset{x\to a}{lim} \frac{g\left(x\right)f\left(a\right)-g\left(a\right)f\left(a\right)+g\left(a\right)f\left(a\right)-g\left(a\right)f\left(x\right)}{x-a}\\ =\underset{x\to a}{lim} f\left(a\right)\left[{\left.\frac{g\left(x\right)-g\left(a\right)}{x-a}\right|}_{\rfloor }-\underset{x\to a}{lim} g\left(a\right)\left[\frac{f\left(x\right)-f\left(a\right)}{x-a}\mid \right]\right.\end{array}$

$\begin{array}{l}=f\left(a\right)\underset{x\to a}{lim} \left[\frac{g\left(x\right)-g\left(a\right)}{x-a}\mid \right]-g\left(a\right)\underset{x\to a}{lim} \left[\frac{f\left(x\right)-f\left(a\right)}{x-a}\mid \right]\\ f\left(a\right)g^{\mathrm{\prime }}\left(a\right)-g\left(a\right)f^{\mathrm{\prime }}\left(a\right)\end{array}$

[by using first principle formula] 

$\begin{array}{l}=3.4-(-1)(-2)=12-2=10\\ \underset{x\to a}{lim} \frac{g\left(x\right)f\left(a\right)-g\left(a\right)f\left(x\right)}{x-a}\end{array}$

Using $L-$Hospital's rule, 

Limit $=\underset{x\to a}{lim} \frac{g^{\mathrm{\prime }}\left(x\right)f\left(a\right)-g\left(a\right)f^{\mathrm{\prime }}\left(x\right)}{1}$

Limit $=g^{\mathrm{\prime }}\left(a\right)f\left(a\right)-g\left(a\right)f^{\mathrm{\prime }}\left(a\right)$

$\begin{array}{l}=\left(4\right)\left(3\right)-(-1)(-2)\\ =12-2=10.\end{array}$